Absolute continuity of the Lebesgue integral

Note that, by the Lebesgue dominated convergence theorem, we have that $$\lim_{\lambda \to\infty}\int_{\{|f| > \lambda\}}|f|\ d\mu = 0.$$ This follows easily since $\chi_{\{|f| > \lambda\}}|f| \le |f| \in L^1$ and $\chi_{\{|f| > \lambda\}}|f| \to 0$ since $f$, being integrable, is finte almost everywhere.

Let $\epsilon > 0$, then there exists $\lambda > 0$ such that $$\int_{\{|f| > \lambda\}}|f|\ d\mu < \frac{\epsilon}{2}.$$

Choose $\delta \le \frac{\epsilon}{2\lambda}$ and take any measurable set $A$ such that $\mu(A) < \delta$. Then we have $$\int_A|f|\ d\mu = \int_{A \cap \{|f| > \lambda\}}|f|\ d\mu + \int_{A \cap \{|f| \le \lambda\}}|f|\ d\mu \le$$ $$\le \int_{\{|f| > \lambda\}}|f|\ d\mu + \int_{A \cap \{|f| \le \lambda\}}\lambda\ d\mu$$

note that this last inequality follows from the fact that $A \cap \{|f| > \lambda\} \subset \{|f| > \lambda\}$ and the fact that $|f| \le \lambda$ on $A \cap \{|f| \le \lambda\}$. Then we are done since $$\int_{\{|f| > \lambda\}}|f|\ d\mu + \int_{A \cap \{|f| \le \lambda\}}\lambda\ d\mu \le \frac{\epsilon}{2} + \delta \lambda \le \epsilon.$$ This concludes the proof! :D

It is given that $f$ is integrable, so $$\int_E|f|d\mu=L<\infty.$$ Choose a simple function $0\leq g\leq|f|$ with $$g=\sum_{i=1}^Ng_i\mathbf{1}_{A_i}$$ such that $$\int_E(|f|-g)d\mu<\frac{\epsilon}{2}.$$

Let $G=\max (g_i)$ and choose $\delta<\frac{\epsilon}{2GN}$.

For any $A$ with $\mu(A)<\delta$ we have $$ \begin{aligned} \int_A|f|d\mu&=\int_A(|f|-g)d\mu+\int_Agd\mu\\ &\leq\int_E(|f|-g)d\mu+\int_Agd\mu\\ &<\frac{\epsilon}{2}+\sum_{i=1}^Ng_i\mu(A_i\cap A)\\ &\leq\frac{\epsilon}{2}+GN\mu(A)\\ &<\frac{\epsilon}{2}+GN\delta\\ &<\frac{\epsilon}{2}+GN\frac{\epsilon}{2GN}=\epsilon. \end{aligned} $$